Lesson Explainer: Graphs of Inverses of Functions Mathematics

In this explainer, we will learn how to use a graph to find the inverse of a function and analyze the graphs for the inverse of a function.

A relation, or mapping, transforms elements from one set onto elements from another. If every input in this mapping has exactly one output, it is called a function.

Definition: Functions

A function maps every element from an input set onto exactly one element from an output set. Functions can be either one-to-one (one input has one output) or many-to-one (many inputs map onto the same output).

If a function 𝑓 maps elements of the sets 𝑋 to π‘Œ, we can use the following notation: π‘“βˆΆπ‘‹βŸΆπ‘Œ.

The set of values that can be input into the function is called the domain, while the set of elements that come out is called the range.

If 𝑓 is a one-to-one function, it is said to be invertible. In other words, there exists an inverse to this function, defined π‘“οŠ±οŠ§ such that the following definition holds.

Definition: Inverse Functions

Let 𝑓 be a function whose domain is the set 𝑋 and whose range is π‘Œ. π‘“οŠ±οŠ§ is the inverse of 𝑓 with domain π‘Œ and range 𝑋 if 𝑓(𝑓(π‘₯))=π‘₯π‘₯𝑋,𝑓𝑓(𝑦)=π‘¦π‘¦π‘Œ.forallinforallin

In other words, the inverse function β€œundoes” the original function. Take, for instance, the function 𝑓(π‘₯)=2π‘₯. 𝑓 takes values of π‘₯ and multiplies them by 2. The inverse of 𝑓 is the function that β€œundoes” this process; hence, 𝑓(π‘₯)=π‘₯2. Note that while there are algebraic processes to calculate the inverse of a function, it is outside the scope of this explainer to explore these in more detail.

By sketching the graph of 𝑦=2π‘₯ and 𝑦=π‘₯2 on the same set of axes, we can identify the single transformation that relates the graph of a function and its inverse.

The graph of 𝑦=2π‘₯ is mapped onto the graph of 𝑦=π‘₯2 by a single reflection in the line 𝑦=π‘₯. This can be generalized for any invertible function 𝑓.

Property: Graph of the Inverse

If 𝑓 is invertible, then the graph of 𝑦=𝑓(π‘₯) is the same as the graph of the equation π‘₯=𝑓(𝑦). This is obtained by reflecting the graph of 𝑦=𝑓(π‘₯) across the line 𝑦=π‘₯.

This is equivalent to switching the roles of π‘₯ and 𝑦 in the function, and so any coordinate for an inverse function can be found by switching the π‘₯- and 𝑦-coordinates of the corresponding point on the graph of the original function. For instance, the point with coordinates (5,10) lies on the graph of 𝑦=2π‘₯, so the image of this point on the graph of 𝑦=π‘₯2 has coordinates (10,5).

In our first example, we will demonstrate how to recognize the graph of an inverse function given the graph of the original function.

Example 1: Identifying the Graph of the Inverse of a Function

The following is the graph of 𝑓(π‘₯)=2π‘₯βˆ’1.

Which one is the graph of the inverse function 𝑓(π‘₯)?

Answer

Recall that, given a function 𝑓, the graph of 𝑦=𝑓(π‘₯) is obtained by reflecting the graph of 𝑦=𝑓(π‘₯) in the line 𝑦=π‘₯. This is equivalent to switching the π‘₯- and 𝑦-coordinates of each point that lies on the line 𝑦=𝑓(π‘₯).

Let’s begin by identifying any three points on the line 𝑦=2π‘₯βˆ’1. We will choose (0,βˆ’1), (1,1), and (2,3).

To find the corresponding coordinates on the graph of the inverse of 𝑓, we switch the π‘₯- and 𝑦-coordinates. The image of the points are therefore (βˆ’1,0), (1,1), and (3,2) respectively. Adding a straight line through these points, we obtain the graph of the inverse, 𝑦=𝑓(π‘₯). We can see from the following sketch that this is a reflection of the graph of 𝑓(π‘₯)=2π‘₯βˆ’1 across the dashed line 𝑦=π‘₯.

The correct answer is (A).

In our previous example, we demonstrated that, by applying the definition of the inverse function, we can relate points on the graph of the function to the image of these points on the graph of the inverse. In our next example, we will perform a similar process on a cubic function.

Example 2: Relating the Graph of a Function to the Graph of Its Inverse Function

Shown is the graph of 𝑓(π‘₯)=5π‘₯+6. Find the intersection of the inverse function 𝑓(π‘₯) with the π‘₯-axis.

Answer

Recall that the graph of an inverse function is found by reflecting the graph of the original function across the line 𝑦=π‘₯. In doing so, the roles of π‘₯ and 𝑦 are interchanged. This means that if the coordinate of the 𝑦-intercept on the graph of 𝑦=𝑓(π‘₯) is (0,π‘Ž) for some real π‘Ž, the image of this point on the graph of 𝑦=𝑓(π‘₯) is (π‘Ž,0). This is the value of an π‘₯-intercept. So, to find the intersection of the inverse function 𝑓(π‘₯) with the π‘₯-axis, we will find the coordinates of 𝑓(π‘₯) with the 𝑦-axis and switch the roles of π‘₯ and 𝑦.

The graph of 𝑦=𝑓(π‘₯) passes through the 𝑦-axis at (0,6). This means that the graph of 𝑦=𝑓(π‘₯) passes through the π‘₯-axis at (6,0).

Let’s demonstrate this graphically. The graph of 𝑓(π‘₯)=5π‘₯+6 is reflected in the dashed line 𝑦=π‘₯ as shown. The π‘₯-intercept is 6, as expected.

In our previous example, we saw that if the coordinate of the 𝑦-intercept on the graph of 𝑦=𝑓(π‘₯) is (0,π‘Ž) for some real π‘Ž, the image of this point on the graph of 𝑦=𝑓(π‘₯) is (π‘Ž,0). It follows that the converse must also be true. Similarly, if the coordinate of the π‘₯-intercept on the graph of 𝑦=𝑓(π‘₯) is (𝑏,0) for some real 𝑏, the image of this point on the graph of 𝑦=𝑓(π‘₯) is (0,𝑏).

In fact, we might even deduce more information about the domain and range of functions and their inverses. Since the outputs of 𝑓 are the inputs of π‘“οŠ±οŠ§, the range of 𝑓 is also the domain of π‘“οŠ±οŠ§. Similarly, since the inputs of 𝑓 are the outputs of π‘“οŠ±οŠ§, the domain of 𝑓 is the range of π‘“οŠ±οŠ§. Further, if a function has no inverse, it might be possible to restrict the domain of that function so that this new function does have an inverse.

For instance, consider the function 𝑓(π‘₯)=π‘₯ whose graph is shown. This function fails the horizontal line test as shown below, so it is not one-to-one.

This means that if we reflect the graph of the function in the line 𝑦=π‘₯, the resulting graph fails the vertical line test; it is a many-to-one mapping and is therefore not the graph of a function. Thus, 𝑓(π‘₯)=π‘₯ does not have an inverse.

However, by restricting the domain of 𝑓 to [0,∞[, the function will pass the horizontal line test and will now be invertible. The graph of its inverse is found by reflecting 𝑦=𝑓(π‘₯) across the line 𝑦=π‘₯, such that the range of π‘“οŠ±οŠ§ is the domain of 𝑓, [0,∞[.

Property: Domain and Range of Inverse Functions

The range of a one-to-one function 𝑓(π‘₯) is the domain of the inverse function 𝑓(π‘₯).

The domain of 𝑓(π‘₯) is the range of 𝑓(π‘₯).

In our next example, we will demonstrate how to apply the relation between a graph of a function and the graph of its inverse to find their points of intersection.

Example 3: Using the Relationship between a Function and Its Inverse to Find Unknowns

The graphs of 𝑓(π‘₯)=π‘₯+π‘οŠ© and its inverse 𝑓(π‘₯) intersect at three points, one of which is ο€Ό45,45.

  1. Determine the value of 𝑏.
  2. Find the π‘₯-coordinate of the point 𝐴 marked on the figure.
  3. Find the π‘₯-coordinate of the point 𝐡 marked on the figure.

Answer

Part 1

Since the graph of 𝑦=𝑓(π‘₯) passes through the point ο€Ό45,45, we can substitute π‘₯=45 and 𝑓(π‘₯)=45 into the equation 𝑓(π‘₯)=π‘₯+π‘οŠ© to find the value of 𝑏: 𝑓(π‘₯)=π‘₯+𝑏45=ο€Ό45+𝑏45=64125+𝑏.

Solving for 𝑏, 𝑏=45βˆ’64125𝑏=36125.

Part 2

The figure shows the graph of 𝑓(π‘₯)=π‘₯+36125 and its inverse. Since the value of the 𝑦-intercept of 𝑦=𝑓(π‘₯) is 36125>0, the red plot is 𝑦=𝑓(π‘₯). This means the blue plot is 𝑦=𝑓(π‘₯). Hence, we can find the coordinates of point 𝐴, which is the π‘₯-intercept of the inverse function, by finding the coordinates of the 𝑦-intercept of the original function and switching π‘₯ and 𝑦. This is equivalent to reflecting this point across the line 𝑦=π‘₯.

The 𝑦-intercept of 𝑓(π‘₯) is ο€Ό0,36125. The π‘₯-intercept of 𝑓(π‘₯) is therefore given by ο€Ό36125,0

Hence, the π‘₯-coordinate of the point 𝐴 is 36125.

Part 3

Recall that the graph of 𝑦=𝑓(π‘₯) is mapped onto 𝑦=𝑓(π‘₯) by a reflection in the line 𝑦=π‘₯. Since point 𝐡 is a point of intersection of 𝑦=𝑓(π‘₯) and 𝑦=𝑓(π‘₯), it must lie on the line 𝑦=π‘₯.

The point of intersection can therefore be found by solving the system of equations 𝑦=π‘₯+36125,𝑦=π‘₯.

Substituting 𝑦=π‘₯ in the first equation and rearranging gives π‘₯=π‘₯+36125π‘₯βˆ’π‘₯+36125=0.

Since one of the points of intersection has a value π‘₯=45, we know that (5π‘₯βˆ’4) is a factor of π‘₯βˆ’π‘₯+36125.

Dividing π‘₯βˆ’π‘₯+36125 by 5π‘₯βˆ’4 gives us the factor 1125ο€Ή25π‘₯+20π‘₯βˆ’9ο…οŠ¨. Hence, our previous equation can be written as 1125(5π‘₯βˆ’4)ο€Ή25π‘₯+20π‘₯βˆ’9=0.

Finally, applying the quadratic formula to the equation 25π‘₯+20π‘₯βˆ’9=0, we find the solutions to be π‘₯=√13βˆ’25,π‘₯=βˆ’βˆš13βˆ’25.

Since the point of intersection lies in the first quadrant, its π‘₯-coordinate must be positive.

Hence, the π‘₯-coordinate of point 𝐡 is π‘₯=√13βˆ’25.

Let’s now demonstrate how to apply the properties of the graphs of inverse functions to sketch a function and its inverse.

Example 4: Finding a Graph of a Function That Is an Inverse of Itself

By sketching the graphs of the following functions, which is the inverse of itself?

  1. 1π‘₯
  2. π‘₯
  3. π‘₯
  4. 1π‘₯

Answer

To answer this question, we will sketch the graph of each function in turn, beginning with the graph of 𝑦=1π‘₯. This is a reciprocal function, with asymptotes at 𝑦=0 and π‘₯=0.

Since finding the inverse is equivalent to switching the roles of π‘₯ and 𝑦 in the function, the asymptotes of the inverse of 1π‘₯ will be π‘₯=0 and 𝑦=0. To sketch the graph of the inverse, we reflect the graph of the original function across the line 𝑦=π‘₯.

This reflection maps the graph of 𝑦=1π‘₯ onto itself, so the answer is A.

We will verify this by checking the other three functions, beginning with π‘₯. π‘₯ is a many-to-one function and is therefore not invertible without performing some restrictions on its domain. Similarly, 1π‘₯ is a many-to-one function and has no inverse.

Next is the function π‘₯. This is a cubic function that passes through the origin.

Reflecting the graph of the function in the line 𝑦=π‘₯ gives the following figure:

Since the graph of the function does not map onto the graph of its inverse, the answer cannot be C.

The correct answer is A, 1π‘₯.

In our final example, we will demonstrate how restricting the domain of a function can make it invertible.

Example 5: Sketching Graphs of Functions to Determine Whether They Are Inverse

By sketching the graphs of 𝑓(π‘₯)=2π‘₯ and 𝑔(π‘₯)=ο„žπ‘₯2 for π‘₯β‰₯0, determine whether they are inverse functions.

Answer

The function 𝑓(π‘₯)=2π‘₯ is a many-to-one functionβ€”that is, a number of inputs to this function will have the same output. This means that it is not an invertible function. However, its domain has been restricted to π‘₯β‰₯0, thereby creating a one-to-one function that does have an inverse.

Since the function 2π‘₯ is quadratic with a positive leading coefficient, it is a U-shaped parabola that passes through the origin. Restricting this to values of π‘₯β‰₯0 gives the following figure.

We will now sketch the graph of 𝑦=ο„žπ‘₯2 on the same axes. This is a transformation of the graph of 𝑦=√π‘₯ by a horizontal stretch of scale factor 2. Its graph is shown below.

It does appear that these functions are inverses of one another, since it looks like they have been reflected across the line 𝑦=π‘₯. We will check by looking at their point of intersection. If this lies on the line 𝑦=π‘₯, its π‘₯- and 𝑦-coordinates will be equal.

To find this point, we will solve 2π‘₯=ο„žπ‘₯2: 4π‘₯=π‘₯28π‘₯βˆ’π‘₯=0π‘₯ο€Ή8π‘₯βˆ’1=0.οŠͺοŠͺ

The solutions to this equation are π‘₯=0 or π‘₯=ο„ž18=12.

Substituting π‘₯=12 into either function gives 𝑦=2Γ—ο€Ό12=12. Since the π‘₯- and 𝑦-values are the same, we know the point of intersection of the curves lie on the line 𝑦=π‘₯.

Similarly, substituting π‘₯=0 into either function gives 𝑦=0. The point (0,0) also lies on the line 𝑦=π‘₯, as demonstrated in the following figure.

We can also check whether these functions are inverses of one another by looking at a couple of points on each line.

The point (1,2) lies on the curve 𝑦=2π‘₯. The image of this point after a reflection across 𝑦=π‘₯ is (2,1). If this lies on the curve 𝑦=ο„žπ‘₯2, substituting π‘₯=2 into this equation will give 𝑦=1: 𝑦=ο„žπ‘₯2=ο„ž22=√1=1.

The image of the point (1,2) after a reflection across 𝑦=π‘₯ lies on the curve 𝑦=ο„žπ‘₯2. Hence, it appears that 𝑓(π‘₯)=2π‘₯ and 𝑔(π‘₯)=ο„žπ‘₯2 for π‘₯β‰₯0 are inverses of one another.

Of course, even though we have looked at a handful of points, this is not quite enough. We could verify this by using graphing software to plot the graphs and their reflections across 𝑦=π‘₯.

In our previous example, we demonstrated how to identify a series of points after a reflection across the line 𝑦=π‘₯ and use this information to determine whether a pair of functions were inverses of one another.

We could verify the result more stringently by using the definition of an inverse function; if 𝑓 be a function whose domain is the set 𝑋 and whose range is π‘Œ, π‘“οŠ±οŠ§ is the inverse of 𝑓 with domain π‘Œ and range 𝑋 if 𝑓(𝑓(π‘₯))=π‘₯π‘₯𝑋,𝑓𝑓(𝑦)=π‘¦π‘¦π‘Œ.forallinforallin

Evaluating the composite function 𝑓𝑔(π‘₯), gives 𝑓𝑔(π‘₯)=π‘“ο€Ώο„žπ‘₯2=2ο€Ώο„žπ‘₯2=2ο€»π‘₯2=π‘₯.

Similarly, 𝑔𝑓(π‘₯)=𝑔2π‘₯=ο„ž2π‘₯2=π‘₯.

They are inverse functions.

Let’s finish by recapping the key concepts from this explainer.

Key Points

  • If 𝑓 is a function whose domain is the set 𝑋 and whose range is π‘Œ, π‘“οŠ±οŠ§ is the inverse of 𝑓 with domain π‘Œ and range 𝑋 if 𝑓(𝑓(π‘₯))=π‘₯π‘₯𝑋,𝑓𝑓(𝑦)=π‘¦π‘¦π‘Œ.forallinforallin
  • If 𝑓 is invertible, then the graph of 𝑦=𝑓(π‘₯) is the same as the graph of the equation π‘₯=𝑓(𝑦). This is obtained by reflecting the graph of 𝑦=𝑓(π‘₯) across the line 𝑦=π‘₯.
  • The range of a one-to-one function 𝑓(π‘₯) is the domain of the inverse function 𝑓(π‘₯), while the domain of 𝑓(π‘₯) is the range of 𝑓(π‘₯).

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